`f(x) = 2csc((3x)/2), (0, 2pi)` Find the points of inflection.

Expert Answers
gsarora17 eNotes educator| Certified Educator




` ` `f'(x)=-3csc(3x/2)cot(3x/2)`

differentiating again,






Now let us find out x for which f''(x)=0,

csc(3x/2)=0 , No value of x exists for which csc(3x/2)=0



So no solution exists.

So,f''(x) is undefined for x=0

Now let us test the concavity of the function by taking test value of -pi/2 and pi/2.







Though the concavity of the function is changing, but that does not mean that the inflection point exists at x=0, since the function is not defined at x=0.

Therefore the inflection points do not exist for the function.